Understanding Electric Charge Distribution in Conducting Spheres

[Phy6dummy]

Homework Statement

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.108 N when their center-to-center separation is 50.0 cm. The spheres are then connected by a thin conducting wire.When the wire is removed, the spheres repel each other with an electrostatic force of 0.0360 N. Of the initial charges on the spheres, with a positive net charge, what was (a) the negative charge on one of them and (b) the positive charge on the other?

Homework Equations

F=k((q1q2)/r^2)

The Attempt at a Solution

What I did was plug and chug
F1=0.108N=8.99^9((q1q2)/0.5^2)=3.0x10^-12
same for the second one F2=.0360 etc.. But I got it wrong.
I can not understand why for the second equation it has to be F2=k((q1+q2)(^2)/4r^2). From what I think, since the charge is conserved, shouldn't it be the same as the original equation? Or am I misinterpreting the book or something? I just can't figure out why Fa=kq1q2/r^2 but Fb=k((q1+q2)(^2)/4r^2) or where did the (q1+q2)/2 came from? Thanks in advance.
Sorry, I am so slow with these things.

 

[jackarms]

How did you come to the value of 3.0 x 10^{-12} for F1? The charges are unknown, so you can't solve it.

And what the second equation means is that the charges become equal after they're connected with the wire, so the charge on each becomes half of the net charge:
$$q = \frac{q_{1} + q_{2}}{2}$$
So the force equation becomes:
$$F = k \cdot \frac{q^{2}}{r^{2}}$$

 

[NascentOxygen]

First, you need to undrstand the implication of connecting both with a piece of wire. What is going on here?

And Welcome to Physics Forums!

 

[Phy6dummy]

And what the second equation means is that the charges become equal after they're connected with the wire, so the charge on each becomes half of the net charge:
$$q = \frac{q_{1} + q_{2}}{2}$$
So the force equation becomes:
$$F = k \cdot \frac{q^{2}}{r^{2}}$$

First, you need to undrstand the implication of connecting both with a piece of wire. What is going on here?

And Welcome to Physics Forums!

Hmm... Can you hold my hand and walk me through this. Lol sorry I am so slow. So, Is it because the charge cannot be created or destroy but can move around the wire means that it is the average of q1 and q2?

And thanks for welcoming me.

 

[NascentOxygen]

The symmetry of the situation means the charges equalize, half on one and the remainder on the other. The copper wire facilitates this rearrangement of the existing charges. It allows charge to move until there is no longer a potential difference between the two spheres.